2022 A/L Physics Past Paper · Question 10 (English Medium)

What the Young's modulus of a wire depends on

The Young's modulus of the material of a wire depends on

(A) the initial length of the wire.

(B) the cross-sectional area of the wire.

(C) the nature of the material of the wire.

Of the above statements,

  1. (1) Only (A) is true.
  2. (2) Only (B) is true.
  3. (3) Only (C) is true.
  4. (4) Only (A) and (C) are true.
  5. (5) All (A), (B) and (C) are true.
Answer and explanation

Answer: (3) Only (C) is true.

What the question asks

We want: which of the statements (A), (B) and (C) are true about what the Young's modulus of a wire depends on.

What each letter means:

YY (N m−2\text{N m}^{-2}): Young's modulus.

FF (N\text{N}): tensile force on the wire.

AA (m2\text{m}^2): area of cross-section of the wire.

ee (m\text{m}): extension of the wire.

ll (m\text{m}): initial length of the wire.

Relevant theory

The ratio of tensile stress to tensile strain is called the Young's modulus of the material

of the wire.

The value of Young's modulus depends not on the dimensions of the sample but on the

nature of its materials.

Reading the options

The options are combinations of (A), (B) and (C).

We test each statement on its own.

For example, think of a long thin rubber band and a short thick rubber band.

The long thin one stretches a lot. The short thick one barely stretches.

But both are made of the same rubber. A steel wire is a different thing again.

So how much a wire stretches depends on its length and its thickness. How stiff the material is does not.

Y=F/Ae/lY = \dfrac{F/A}{e/l}

In words: the Young's modulus is the stress divided by the strain.

A longer wire stretches more, but its ee and its ll grow together, so el\dfrac{e}{l} is the same. A thicker wire needs a bigger FF, but FF and AA grow together, so FA\dfrac{F}{A} is the same.

Reasoning and calculation

the initial length of the wire.

Make the wire twice as long. For the same pull the extension ee doubles too, so el\dfrac{e}{l} is unchanged. So YY is unchanged.

A

the cross-sectional area of the wire.

Make the wire twice as thick. The force FF for the same stress doubles too, so FA\dfrac{F}{A} is unchanged. So YY is unchanged.

B

the nature of the material of the wire.

Steel, copper and brass give different values of YY, even for wires of the same size. YY depends on the nature of the material.

C

Statement (A) is false, because the length is divided out in the strain.

Statement (B) is false, because the area is divided out in the stress.

Statement (C) is true.

So only (C) is true.

Why the other options fail

Statement (A) is false, because the length is divided out. Statement (C) is true.

Statement (B) is false, because the area is divided out. Statement (C) is true.

Statement (A) is false, because the length is a dimension of the sample, not of the material.

Statements (A) and (B) are false, because the length and the area are dimensions of the sample.

Conclusion

Only statement (C) is true. (A) and (B) are false.

The answer is (3).

The extension of a wire depends on its length, its thickness and its material. The Young's modulus depends on the material only.

Length and area appear in Y=FlAeY = \dfrac{F l}{A e}, but they cancel.

Exam tip

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